Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{1}{{{{\cos }^2}x - {{\sin }^2}x}} + \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}\\
= \dfrac{1}{{\cos 2x}} + \dfrac{{2\dfrac{{\sin x}}{{\cos x}}}}{{1 - \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}}\\
= \dfrac{1}{{\cos 2x}} + \dfrac{{2\sin x}}{{\cos x}}:\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x}}\\
= \dfrac{1}{{\cos 2x}} + \dfrac{{2\sin x}}{{\cos x}}.\dfrac{{{{\cos }^2}x}}{{\cos 2x}}\\
= \dfrac{1}{{\cos 2x}} + \dfrac{{2\sin x.\cos x}}{{\cos 2x}}\\
= \dfrac{{1 + 2\sin x.\cos x}}{{\cos 2x}}\\
= \dfrac{{1 + \sin 2x}}{{\cos 2x}}\\
= \dfrac{{\left( {1 - \sin 2x} \right)\left( {1 + \sin 2x} \right)}}{{\cos 2x\left( {1 - \sin 2x} \right)}}\\
= \dfrac{{1 - {{\sin }^2}2x}}{{\cos 2x.\left( {1 - \sin 2x} \right)}}\\
= \dfrac{{{{\cos }^2}2x}}{{\cos 2x.\left( {1 - \sin 2x} \right)}}\\
= \dfrac{{\cos 2x}}{{1 - \sin 2x}}
\end{array}\)
